JEE Main & Advanced Physics Simple Harmonic Motion JEE PYQ-Simple Harmonic Motion

  • question_answer
    The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be- [JEE Main 11-Jan-2019 Evening]

    A) \[2\sqrt{3}s\]     

    B)      \[\frac{3}{2}s\]

    C) \[\frac{2}{\sqrt{3}}s\]    

    D)                  \[\frac{\sqrt{3}}{2}s\]

    Correct Answer: A

    Solution :

    [a] Time period of a simple pendulum,
    \[T=2\pi \sqrt{\frac{l}{g}}\]
    \[{{g}_{p}}=\frac{G{{M}_{p}}}{R_{p}^{2}}=4\frac{G{{M}_{p}}}{D_{p}^{2}}=\frac{(4G{{M}_{E}})}{D_{E}^{2}}\times \frac{3}{{{3}^{2}}}\]
    \[\left( \because {{g}_{e}}=\frac{4G{{M}_{E}}}{D_{E}^{2}} \right)\]
    \[{{g}_{p}}=\frac{{{g}_{e}}}{3}\]
    \[\therefore \]\[\frac{{{T}_{e}}}{{{T}_{p}}}=\sqrt{\frac{{{g}_{p}}}{{{g}_{e}}}}=\frac{1}{\sqrt{3}},{{T}_{p}}=2\sqrt{3}s\]            \[({{T}_{e}}=2s)\]


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